A population is normally distributed with µ = 200 and o = 10.a. Find the probability that a value randomly selected from this population will have a value greater than 215.b. Find the probability that a value randomly selected from this population will have a value less than 190.c. Find the probability that a value randomly selected from this population will have a value between 190 and 215.Click the icon to view the standard normal table.a. P(x>215) =(Round to four decimal places as needed.)b. P(x < 190) =(Round to four decimal places as needed.)c. P(190

From the provided information, Mean (µ) = 200 Standard deviation (σ) = 10

Answered: A population is normally distributed…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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For the standard normal distribution, find the z-scores that enclose the middle 92% of the data. a. (-1.34, 1.34) b. (-2.03, 2.03) c. (-1.75, 1.75) d. (-1.04, 1.04)
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Use the standard normal table to find the z-score that corresponds to the given percentile. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score. P18 The z-score that corresponds to P18 is
Use the standard normal table to find the z-score that corresponds to the given percentile. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score. P 25 Click to view page 1 of the table. Click to view page 2 of the table. The z-score that corresponds to P25 is (Round to two decimal places as needed.)
Use the standard normal table to find the z-score that corresponds to the given percentile. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score. P7 The z-score that corresponds to P7 is
Use the standard normal table to find the z-score that corresponds to the given percentile. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score. P 48 Click to view page 1 of the table. Click to view page 2 of the table. The Z-score that corresponds to P48 is (Round to two decimal places as needed.)
K The lengths of pregnancies are normally distributed with a mean of 269 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 308 days or longer. b. If the length of pregnancy is in the lowest 3%, then the baby is premature. Find the length that separates premature babies from those who are not premature. Click to view page 1 of the table. Click to view page 2 of the table. a. The probability that a pregnancy will last 308 days or longer is (Round to four decimal places as needed.) b. Babies who are born on or before (Round to the nearest integer as needed.) days are considered premature. .
Refer to the data set in the accompanying table. Assume that the paired sample data is a simple random sample and the differences have a distribution that is approximately normal. Use a significance level of 0.05 to test for a difference between the weights of discarded paper (in pounds) and weights of discarded plastic (in pounds). LOADING... Click the icon to view the data. In this example, μd is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the weight of discarded paper minus the weight of discarded plastic for a household. What are the null and alternative hypotheses for the hypothesis test? A. H0: μd≠0 H1: μd>0 B. H0: μd=0 H1: μd<0 C. H0: μd=0 H1: μd≠0 D. H0: μd≠0 H1: μd=0 Household Paper Plastic 1 6.44 8.40 2 7.72 3.86 3 3.27 0.63 4 9.19 3.74 5 14.33 6.43 6 11.08 12.47 7…
The average number of kilos of meat a person consumes in a year is 100 kilos. Assume that the standard deviation is 11 kilos and the distribution is approximately normal. If a sample of 40 individuals is selected, find the probability that the mean of the sample will be 105 kilos or greater per year. Express your answer in percent form with 2 decimal places (with the percent symbol - e.g. 10.00%). type your answer... Next
Assume that adults have IQ scores that are normally distributed with a mean of u = 105 and a standard deviation o = 15. Find the probability that a randomly selected adult has an IQ between 89 and 121. Click to view page 1 of the table. Click to view page 2 of the table. Standard Normal Table (Page 2) The probability that a randomly selected adult has an IQ between 89 and 121 is (Type an integer or decimal rounded to four decimal places as needed.) POSITIVE z Scores Standard Normal (z) Distribution: Cumulative Area from the LEFT .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 5279 .5319 .5359 0.1 .5398 .5438 5478 .5517 .5557 .5596 ,5636 ,5675 .5714 .5753 0.2 .5793 .5832 5871 5910 .5948 .5987 .6026 .6064 .6103 .6141 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 0.5 .6915 6950 .6985 7019 .7054 .7088 7123 7157 .7190 .7224 0.6 .7257 .7291 7324 7357 7389 7422 7454 7486…
Use Table A (Standard Normal Distribution) to find the proportion of observations that satisfies each of the following statements. In each case, shade the area under the curve that is the answer to the question. The table chart below is the one that needs to be looked at to find the proportion of observations that satisfies each of the following statements. The table chart along with the questions that needs to be correctly answered and shaded are below. This deals with Measures of Variations.

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